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Problem: Suppose $f$ is analytic on the domain $\Omega$ except at the isolated singularity $a \in \Omega$. Show that $a$ is a removable singularity if $\mathrm{Im}(f(z))$ is bounded from above.

Attempt:

  1. We have that $a$ is a removable singularity if and only if we have that

    $$ \lim_{z\rightarrow a} (z-a)f(z) =0 $$

  2. Since $\mathrm{Im}(f(z))$ is bounded from above, there exists some $M \in \mathbb{R}_{\ge 0}$ s.t. $\mathrm{Im}(f(z)) \le M$ for all $z \in \Omega$.

  3. If we could show

    $$ \mathrm{Im}(f(z)) \le M \implies |f(z)| \le M_1 \text{ for some } M_1 \in \mathbb{R}_{\ge 0} $$

    then we would immediately have (1) since

    $$ \lim_{z \rightarrow a}|(z-a)f(z)| \le \lim_{z \rightarrow a}|(z-a)|M_1 = 0 $$

Question: Is this the right approach?

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    $\begingroup$ Your point 3. is not very clear; rather consider the function $1/(f(z)-i(M+1))$, which is bounded $\endgroup$ – user8268 Apr 22 '14 at 11:49
  • $\begingroup$ I think the approach idea is correct, but I'm not sure whether it is feasible. $\endgroup$ – DonAntonio Apr 22 '14 at 11:55
  • $\begingroup$ @user8268: I see that $g(z) = 1 / (f(z) - i(M+1))$ is analytic on $\Omega$. I see that if $g(z)$ weren't bounded, then it would have to be that $f(z)$ were bounded (whence the theorem would follow). So I assume we'd be trying to find a contradiction if $g(z)$ is assumed to be bounded. The only thing to contradict would be the analyticity of $f$ or the fact that $\mathrm{Im}(f(z))$ is bounded from above. I'm not sure how assuming $g(z)$ is bounded contradicts those facts though. $\endgroup$ – user1770201 Apr 22 '14 at 13:07
  • $\begingroup$ Can you use Picard's theorem? $\endgroup$ – Najib Idrissi Apr 29 '14 at 7:37
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Update:

Assume that ${\rm Im}\bigl(f(z)\bigr)<M$ for some $M>0$ and all $z\in\dot\Omega:=\Omega\setminus\{a\}$. The Moebius transform $$w\mapsto \zeta:={iMw\over 2iM-w}$$ maps the half plane ${\rm Im}(w)<M$ onto the interior of the disc $|\zeta|<M$. It follows that the function $$g(z):={iM f(z)\over 2iM-f(z)}$$ satisfies $$|g(z)|<M\qquad\forall z\in\dot\Omega\ .$$ Therefore $g$ has a removable singularity at $a$ and is in fact analytic in all of $\Omega$. By the maximum principle it follows that $|g(a)|<M$ as well, whence $$f(z)={2iM g(z)\over iM+g(z)}$$ is analytic in all of $\Omega$.

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  • $\begingroup$ What is meant by $o(1)$? $\endgroup$ – user1770201 Apr 29 '14 at 11:11
  • $\begingroup$ As it stands, it's not obvious to me how adjusting $\phi$ can make the imaginary part of $f$ arbitrarily large. It seems like it will still be bounded by $\left|{1 \over r^n}(c + o(1))\right|$. $\endgroup$ – user1770201 Apr 29 '14 at 11:41
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The function $g:=-if$ satisfies $Re(g)=Im(f)$, so $Re(g)$ is bounded above near $a$; and of course it is enough to show that $g$ has a removable singularity at $a$.

Choose $r>0$ such that $Re(g)$ is bouded above on $D^*(a,r):=D(a,r)\setminus\{ a\}$. Consider the function $\phi=e^g$. This function is holomorphic in $D^*(a,r)$ and bounded therein since $\vert \phi\vert=e^{Re(g)}$. Hence, $\phi$ has a removable singularity at $a$. Let us still denote by $\phi$ the analytic continuation of $\phi$ defined on $D(a,r)$.

Write $\phi(z)=(z-a)^k\psi (z)$, where $k$ is a nonnegative integer and $\psi$ is holomorphic in $D(a,r)$ with $\psi(a)\neq 0$. Then $\psi$ has no zeros in $D(a,r)$ because $\phi=e^g$ has no zeros in $D^*(a,r)$. So one can find a function $h$ holomorphic in $D(a,r)$ such that $e^{h(z)}=\psi(z)$ in $D(a,r)$. Then $\phi(z)=(z-a)^ke^{h(z)}$, so that $$\forall z\in D^*(a,r)\;:\; e^{g(z)}=(z-a)^k e^{h(z)}\, .$$ In particular, the function $u_k(z)= (z-a)^k$ has a holomorphic logarithm on $D^*(a,r)$, namely $l(z)=g(z)-h(z)$. This forces $k=0$. Indeed, if $C\subset D^*(a,r)$ is any (positively oriented) circle with center $a$, then on the one one hand the winding number of $u_k(C)$ around $a$ must be equal to $0$ since $u_k$ has a holomorphic logarithm, and on the other hand this winding number is equal to $k$. "Alternatively", observe that $$2i\pi k= \int_{\partial D(a,r/2)} \frac{k}{z-a}\, dz=\int_{\partial D(a,r/2)} l'(z)\, dz=0\, .$$ So we obtain that $e^{g(z)-h(z)}\equiv 1$ on $D^*(a,r)$. This means that $g(z)-h(z)\in2i\pi \mathbb Z$ for all $z\in D^*(a,r)$, and since $D^*(a,r)$ is connected, it follows that there exists an integer $n\in\mathbb Z$ such that $$\forall z\in D^*(a,r)\;:\; g(z)-h(z)=2i\pi n\, .$$ Hence, the function $h+2i\pi n$ is an analytic continuation of $g$ in the disk $D(a,r)$, which gives the required conclusion.

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Well, as you say, you certainly have $$\begin{eqnarray} 1. && |f| \text{ bounded} \Rightarrow \lim_{z\to a} (z-a)f(z) = 0 \\ 2. && |f| \text{ bounded} \Rightarrow \textrm{Im }f \text{ bounded.} \end{eqnarray}$$

To show that $\lim_{z\to a} (z-a)f(z) = 0$ implies $\textrm{Im } f$ is bounded, you'd need the reverse direction of (1), and to shows that $\textrm{Im } f$ is bounded implies $\lim_{z\to a} (z-a)f(z) = 0$ the reverse direction of (2).

Both reverse directions will need to use that $f$ is analytic, because they aren't true otherwise. You can get the reverse direction of (1) by looking the the laurent series of $f(z)$ around $a$, I think. Dunno how you'd get the reverse direction of (2) - though maybe looking at the laurent series of $f$ yields something again, if you choose your $x$ carefully...

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Holomorphic functions are very very firms. So I thought: there must be a link (beyond Cauchy-Riemann conditions) between the behaviour of the real and the imaginary part of a such function.

Let $ f:\Omega\setminus\{a\}\longrightarrow\mathbb C $ be an holomorphic function where $a$ is an isolated singularity and $\exists M\ge0$ s.t. $|v|\le M$ on $\Omega\setminus\{a\}$, with $f=u+iv$.

Then $a$ is a removable singularity.

Let's show that if $a$ is not a removable singularity then $v$ can't be bounded on $\Omega\setminus\{a\}$.

There are only two possibilities for $a$ to be not a removable singularity: be a pole or an essential singularity.


Case 1: $a$ is a pole.

Let's write the Laurent series for $f$ around $a$, pole of order $m$: $$ f(z)=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n}+ \sum_{n=0}^{+\infty}c_{n}(z-a)^n\;,\;\;\;z\in B(a,r[\setminus\{a\} $$ where $$ P_{f,a}(z)=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n} $$ is the principal (or sometimes singular) part of $f$ around $a$. The name "principal part" is because the nature of the singularity is its her responsability.

Now, $a$ is a pole for $f$ iff $\lim_{z\rightarrow a}f(z)=\infty_{\mathbb C}$, and naturally this holds iff $\lim_{z\rightarrow a}P_{f,a}(z)=\infty_{\mathbb C}$.

Then, being $v$ bounded, $\lim_{z\rightarrow a}f(z)=\infty_{\mathbb C}$ iff $\lim_{z\rightarrow a}u(z)=\infty$ which implies that $\lim_{z\rightarrow a}\Re(P_{f,a}(z))=\infty$. Moreover $v$ bounded implies $\Im P_{f,a}$ bounded.

Let's now reach the contradiction showing by induction on $m$ that $$ \left. \begin{array}{lllllll} (a) \; \lim_{z\rightarrow a}P_{f,a}(z)=\infty_{\mathbb C}\\ (b) \; \lim_{z\rightarrow a}\Re(P_{f,a}(z))=\infty\\ \end{array} \right\}\Longrightarrow \lim_{z\rightarrow a}\Im(P_{f,a}(z))=\infty\;. $$

Let us suppose now $m=1,\; z=x+iy,\; a=\alpha+i\beta$.

\begin{align*} P_{f,a}(z)=&\frac{c_{-1}}{z-a}\\ =&\frac{\Re c_{-1}+i\Im c_{-1}}{(z-a)(\overline{z-a})}\overline{z-a}\\ =&\frac{\Re c_{-1}(x-\alpha)+\Im c_{-1}(y-\beta)}{(x-\alpha)^2+(y-\beta)^2}+ i\frac{\Im c_{-1}(x-\alpha)-\Re c_{-1}(y-\beta)}{(x-\alpha)^2+(y-\beta)^2} \end{align*} Then\begin{align*} \Re(P_{f,a}(z))=&\frac{\Re c_{-1}(x-\alpha)+\Im c_{-1}(y-\beta)}{(x-\alpha)^2+(y-\beta)^2}\\ =&\frac{\Re c_{-1}}{(x-\alpha)+\frac{(y-\beta)^2}{(x-\alpha)}}+ \frac{\Im c_{-1}}{(y-\beta)+\frac{(x-\alpha)^2}{(y-\beta)}}\;; \end{align*} and being by hypotesis $\lim_{z\rightarrow a}\Re(P_{f,a}(z))=\infty$, then at least one between $\frac{(x-\alpha)^2}{(y-\beta)}$ and $\frac{(y-\beta)^2}{(x-\alpha)}$ must tend to $0$ as $z\rightarrow a$ (equivalently $(x,y)\rightarrow(\alpha,\beta)$). Then we can conclude, with a simmetry argument, that also $\lim_{z\rightarrow a}\Im(P_{f,a}(z))=\infty$.

Note that the role of $c_{-1}$ is not trivial: we have supposed $\Re c_{-1}\neq0\neq\Im c_{-1}$; the other two cases are easier; if $\Re c_{-1}\neq0=\Im c_{-1}$, suppose wlog $c_{-1}=1$ and $a=0$. Then we're facing the case $P_{f,a}(z)=\frac1{z}$. Now $\frac1{z}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$; by hypotesis $\frac{x}{x^2+y^2}\rightarrow\infty$ as $(x,y)\rightarrow(0,0)$. But $a=0\in\Omega$ hence $\exists t>0$ s.t. $B(0,t[\subseteq\Omega$, so we can argue by simmetry to conclude that also $\frac{y}{x^2+y^2}\rightarrow\infty$ as $(x,y)\rightarrow(0,0)$.

The case $\Re c_{-1}=0\neq\Im c_{-1}$ is similar.

Inductive step: call now $$ P_{f,a}^{m}=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n} $$ (we only highlight the $m$). So, let the statement true for $P_{f,a}^{m-1}$ and show it holds for $P_{f,a}^{m}$. $$ P_{f,a}^{m} =\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n} =\frac{c_{-m}}{(z-a)^m}+P_{f,a}^{m-1} $$ so $$ \Re\left(P_{f,a}^{m}\right) =\Re\left(\frac{c_{-m}}{(z-a)^m}\right)+\Re\left(P_{f,a}^{m-1}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty $$ hence exists at least one $k$ in $\{1,\dots,m\}$ s.t. $\Re\left(\frac{c_{-k}}{(z-a)^k}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$:

$\bullet$ if $k\le m-1$ then $\Re\left(P_{f,a}^{m-1}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$ and by inductive hyp we have $\Im\left(P_{f,a}^{m-1}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$ and so also $\Im\left(P_{f,a}^{m}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$ which concludes.

$\bullet$ if $k=m$, observing that $ \frac{c_{-m}}{(z-a)^m}=\frac{c_{-m}}{(z-a)^{m-1}}\frac1{z-a}\; $ we have \begin{align*} \Re\left(\frac{c_{-m}}{(z-a)^m}\right)=& \Re\left(\frac{c_{-m}}{(z-a)^{m-1}}\frac{1}{z-a}\right)\\ =&\underbrace{\Re\left(\frac{c_{-m}}{(z-a)^{m-1}}\right)}_{\xi_1} \underbrace{\Re\left(\frac1{(z-a)}\right)}_{\xi_2}- \underbrace{\Im\left(\frac{c_{-m}}{(z-a)^{m-1}}\right)}_{\eta_1} \underbrace{\Im\left(\frac1{(z-a)}\right)}_{\eta_2} \end{align*} which goes to infinity; hence al least one between $\xi_1,\xi_2,\eta_1,\eta_2$ goes to infinity; if were $\xi_{1}$ and/or $\xi_2$ we fall in the previous case then we conclude.

If otherwise were one of the $\eta$'s then $\Im\left(P_{f,a}^{m}\right)\rightarrow\infty$, so we have finished too.

Then we reached the contradiction , so $a$ can't be a pole.


Case 2: $a$ is an essential singularity.

Simply use Casorati-Weierstra$\beta$ theorem: let $U$ be an open nhb of $a$ in $\Omega$. Then $f(U\setminus\{a\})$ is dense in $\mathbb C$, hence $v$ can't be bounded, absurd.


So we proved that the isolated singularity can't be neither a pole nor an essential singularity. Hence it must be a removable singularity.

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See [Casorati–Weierstrass theorem][1]

If $g(z)$ is bounded for $0<|z-a|<r$ then $g$ has a removable singularity at $z=a$. This is a simple corollary of Cauchy's formula.

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